Find the inverse of the matrix, $\text B = \left[\begin{array}{rr}-4 & 2 \\ -6 & 6\end{array}\right]$. Non-integers should be given either as decimals or as simplified fractions. $ B^{-1}=$
Solution: The Strategy To find the inverse of an invertible matrix, we can use Gaussian Elimination. To do this, we do the following. First, we append the matrix $\text B$ with the identity matrix $\text I$ to get [ B | I ] \left[\begin{array} ~\text B ~ |~\text I\end{array}\right]. Next, we use Gaussian Elimination to reduce $\text B$ to the identity matrix, $\text I$. Performing the same operations on $\text I$ will convert it to $\text B^{-1}$, so that our new matrix becomes [ I | B − 1 ] \left[\begin{array} ~\text I ~ |~\text B^{-1}\end{array}\right]. Appending $\text B$ with $\text I$ [ B | I ] = [ − 4 − 6 2 6 1 0 0 1 ] \left[\begin{array} ~\text B ~ |~\text I\end{array}\right]=\left[\begin{array}{rr}-4 & 2 & 1 & 0 \\ -6 & 6 & 0 & 1 \end{array}\right] Eliminating the leading term in the second row We want the first term of $R_2$ to equal $0$, so we subtract $\dfrac{3}{2}R_1$ from $R_2$. $\left[\begin{array}{rr}-4 & 2 & 1 & 0 \\ {-6} & {6} & {0} & {1} \end{array}\right]\xrightarrow{R_2-\dfrac{3}{2}R_1\rightarrow R_2}\left[\begin{array}{rr}-4 & 2 & 1 & 0 \\ {0} & {3} & {-\dfrac{3}{2}} & {1} \end{array}\right]$ Reducing the leading terms and back-solving Now, let's reduce the leading term of $R_2$ to equal $1$. $\left[\begin{array}{rr}-4 & 2 & 1 & 0 \\ {0} & {3} & {-\dfrac{3}{2}} & {1} \end{array}\right]\xrightarrow{\dfrac{1}{3}R_2\rightarrow R_2}\left[\begin{array}{rr}-4 & 2 & 1 & 0 \\ {0} & {1} & {-\dfrac{1}{2}} & {\dfrac{1}{3}} \end{array}\right]$ We are ready to back-solve to get [ I | B − 1 ] \left[\begin{array} ~\text I ~ |~\text B^{-1}\end{array}\right]. $\begin{aligned}\!\!\left[\begin{array}{rr}\!{-4}\! & {2}\! & \!\!\!{1}\! & {0} \\ \!0\! & 1\! & \!\!-\!\dfrac{1}{2}\! &\!\! \dfrac{1}{3} \!\end{array}\right]\!\xrightarrow{\!\!R_1-2R_2\rightarrow R_1\!\!} \!\!&\left[\begin{array}{rr}\!{-4}\!\! & {0}\!\! & \!\!\!{2} \!\!& {-\dfrac{2}{3}} \!\\ \!0\! & 1\! & \!\!-\!\dfrac{1}{2}\! &\!\!\! \dfrac{1}{3} \end{array}\right] \!\!\xrightarrow{\!\!-\!\dfrac{1}{4}R_1\rightarrow R_1\!\!}\!\!\left[\begin{array}{rr}{1}\!\! & {0}\!\! & {-\dfrac{1}{2}}\!\! & {\dfrac{1}{6}} \\ \!0\! & 1\! & \!\!-\dfrac{1}{2}\! &\!\! \dfrac{1}{3} \end{array}\right]\end{aligned}$ Therefore $\text B^{-1}=\left[\begin{array}{rr} -\dfrac{1}{2}\!\! & \dfrac{1}{6} \\ -\dfrac{1}{2} \!\!& \dfrac{1}{3} \end{array}\right]$. Summary $\text B^{-1}=\left[\begin{array}{rr} -\dfrac{1}{2}\!\! & \dfrac{1}{6} \\ -\dfrac{1}{2} \!\!& \dfrac{1}{3} \end{array}\right]$